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hdu 1565&&hdu 1569 (最大点权独立集)

2018年02月20日 ⁄ 综合 ⁄ 共 1324字 ⁄ 字号 评论关闭

题目意思很明确就是选一些没有相连的数字,使和最大,建成二分图后求最大点权独立集,,





#include<stdio.h>
#include<string.h>
const int N=2510;
const int inf=0x3fffffff;
int dis[N],gap[N],head[N],num,start,end,ans;
struct edge
{
	int ed,flow,next;
}e[N*6];
void addedge(int x,int y,int w)
{
	e[num].ed=y;e[num].flow=w;e[num].next=head[x];head[x]=num++;
	e[num].ed=x;e[num].flow=0;e[num].next=head[y];head[y]=num++;
}
int dfs(int u,int minflow)
{
	if(u==end)return minflow;
	int i,v,f,min_dis=ans-1,flow=0;
	for(i=head[u];i!=-1;i=e[i].next)
	{
		v=e[i].ed;
		if(e[i].flow<=0)continue;
		if(dis[v]+1==dis[u])
		{
			f=dfs(v,e[i].flow>minflow-flow?minflow-flow:e[i].flow);
			e[i].flow-=f;
			e[i^1].flow+=f;
			flow+=f;
			if(flow==minflow)break;
			if(dis[start]>=ans)return flow;
		}
		min_dis=min_dis>dis[v]?dis[v]:min_dis;
	}
	if(flow==0)
	{
		if(--gap[dis[u]]==0)
			dis[start]=ans;
		dis[u]=min_dis+1;
		gap[dis[u]]++;
	}
	return flow;
}
int isap()
{
	int maxflow=0;
	memset(dis,0,sizeof(dis));
	memset(gap,0,sizeof(gap));
	gap[0]=ans;
	while(dis[start]<ans)
		maxflow+=dfs(start,inf);
	return maxflow;
}
int main()
{
	int i,n,w,j,x,sum,m;
	while(scanf("%d%d",&n,&m)!=-1)
	{
		memset(head,-1,sizeof(head));
		num=0;sum=0;start=0;end=n*m+1;ans=end+1;
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				x=i*m+j-m;
				scanf("%d",&w);
				sum+=w;
				if((i+j)%2!=0)
					addedge(x,end,w);
				else 
				{
					addedge(start,x,w);
					if(i>1)addedge(x,x-m,inf);
					if(j>1)addedge(x,x-1,inf);
					if(i<n)addedge(x,x+m,inf);
					if(j<m)addedge(x,x+1,inf);
				}
			}
		}
		printf("%d\n",sum-isap());
	}
	return 0;
}

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